Block A of mass `m//2` is connected to one end of light rope which passes over a pulley as shown in fig. A man of mass m climbs the other end of rope with a relative acceleration of `g//6` with respect to rope. Find the acceleration of block A and tension in the rope.

Explanation

Let the acceleration of block be `a_(0)` up. The acceleration of man with respect to ground is a .
`:. g/6=a+a_(0) implies a=g/6-a_(0)`
From the force diagram of man,
`T=mg=ma=m((g)/(6)-a_(0))` ..(i)
From force diagram of block A,
`T-(m)/(2)g=(m)/(2)a_(0)`...(ii)
Substracting (i) from (ii)
`(m)/(2) g = (m)/(2) a_(0)-(mg)/(6)+ma_(0)`
or, `(3)/(2) ma_(0)=(mg)/(2)+(mg)/(6) implies a_(0)=(4)/(9)g`
From (ii), `T=m/2 a_(0)+m/2 g`
`=m/2 xx 4/9 g +m/2 g implies T =(13mg)/(18)`.